As only the outer surface of the sphere is charged, consider the element as a ring, as shown in the figure. The equivalent current due to the ring element,
$d i=\frac{\omega}{2 \pi}(2 \pi r \sin \theta n d \theta) \sigma$
and magnetic induction due to this loop element at the centre of the sphere, $O$, $d B=\frac{\mu_{0}}{4 \pi} d i \frac{2 \pi r \sin \theta r \sin \theta}{r^{3}}=\frac{\mu_{0}}{4 \pi} d i \frac{\sin ^{2} \theta}{r}$
[Using $3.219$ (b) ] Hence, the total magnetic induction due to the sphere at the centre, $O$, $\pi / 2$
$B=\int d B=\int_{0} \frac{\mu_{0}}{4 \pi} \frac{\omega}{2 \pi} \frac{2 \pi r^{2} \sin \theta d \theta \sin ^{2} \theta \sigma}{r}$ [using (1)]
$\pi / 2$
Hence, $\quad B=\int_{0} \frac{\mu_{0} \sigma \omega r}{4 \pi} \sin ^{3} \theta d \theta=\frac{2}{3} \mu_{0} \sigma \omega r=29 \mathrm{pT}$